Chapter 9, Problem 25CT

To determine

Calculate the polar equation of an hyperbola.

Answer to Problem 25CT

The polar equation is $r=\frac{10}{4+5\sin \theta }$ .

Explanation of Solution

Given: It is given in the question that the focus of the hyperbola is at the pole, an eccentricity is $e=\frac{5}{4}$ , and directrix at $y=2$ .

Concept Used:

In this, use the concept of understanding the concept of hyperbola and also use the general equation of hyperbola $r=\frac{ep}{1+e\sin \theta }$ .

Calculation: Here, the distance between the pole and the directrix is p,

i.e, $p=\left|2\right|=2$

Since, the directrix is horizontal and is above the pole. Use e and p to write the equation of the hyperbola,

So, $e=\frac{5}{4}$

  $p=2$ ,then use the general equation

  $\begin{array}{l}r=\frac{ep}{1+e\sin \theta }\\ r=\frac{\frac{5}{4}\times 2}{1+\frac{5}{4}\sin \theta }\\ r=\frac{\frac{5}{2}}{1+\frac{5}{4}\sin \theta }\\ r=\frac{\frac{5}{2}\times 4}{4+5\sin \theta }\\ r=\frac{10}{4+5\sin \theta }\end{array}$

Conclusion:

The equation is $r=\frac{10}{4+5\sin \theta }$ .